Difference between revisions of "2021 AIME II Problems/Problem 7"
MRENTHUSIASM (talk | contribs) (→Solution 4 (No Casework: Two Variables, Two Equations)) |
MRENTHUSIASM (talk | contribs) (→Solution 4 (No Casework: Two Variables, Two Equations)) |
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==Solution 4 (No Casework: Two Variables, Two Equations)== | ==Solution 4 (No Casework: Two Variables, Two Equations)== | ||
− | Number the given equations <math>(1),(2),(3),</math> and <math>(4),</math> in this order. Rearranging <math>(2)</math> | + | Number the given equations <math>(1),(2),(3),</math> and <math>(4),</math> in this order. Rearranging <math>(2)</math> and solving for <math>c,</math> we have |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
ab+(a+b)c&=-4 \\ | ab+(a+b)c&=-4 \\ | ||
ab-3c&=-4 \\ | ab-3c&=-4 \\ | ||
− | c&=\frac{ab+4}{3}. \hspace{ | + | c&=\frac{ab+4}{3}. \hspace{14mm} (5) |
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | Substituting <math>(5)</math> into <math>(4)</math> and solving for <math>d,</math> we get | ||
+ | <cmath>\begin{align*} | ||
+ | ab\left(\frac{ab+4}{3}\right)d&=30 \\ | ||
+ | d&=\frac{90}{ab(ab+4)}. \hspace{5mm} (6) | ||
+ | \end{align*}</cmath> | ||
+ | Substituting <math>(5)</math> and <math>(6)</math> into <math>(3)</math> and simplifying, we rewrite the left side of <math>(3)</math> in terms of <math>a</math> and <math>b</math> only: | ||
<b>SOLUTION IN PROGRESS. NO EDIT PLEASE--A MILLION THANKS!</b> | <b>SOLUTION IN PROGRESS. NO EDIT PLEASE--A MILLION THANKS!</b> |
Revision as of 02:17, 28 March 2021
Contents
Problem
Let and
be real numbers that satisfy the system of equations
There exist relatively prime positive integers
and
such that
Find
.
Solution 1
From the fourth equation we get substitute this into the third equation and you get
. Hence
. Solving we get
or
. From the first and second equation we get
, if
, substituting we get
. If you try solving this you see that this does not have real solutions in
, so
must be
. So
. Since
,
or
. If
, then the system
and
does not give you real solutions. So
. Since you already know
and
, so you can solve for
and
pretty easily and see that
. So the answer is
.
~ math31415926535
Solution 2 (Easy Algebra)
We can factor out of the last two equations. Therefore, it becomes
. Notice this is just
, since
. We now have
and
.
We then find
in terms of
, so
. We solve for
and find that it is either
or
. We can now try for these two values, and plug the rest into the equation. Thus, we have
. We have
and we're done.
~Arcticturn
Solution 3 (Easy Algebra)
can be rewritten as
.
Hence,
Rewriting , we get
.
Substitute
and solving, we get,
call this Equation 1
gives
.
So,
, which implies
or
call this equation 2.
Substituting Eq 2 in Eq 1 gives,
Solving this quadratic yields that
Now we just try these 2 cases.
For substituting in Equation 1 gives a quadratic in
which has roots
Again trying cases, by letting , we get
, Hence
We know that
, Solving these we get
or
(doesn't matter due to symmetry in a,b)
So, this case yields solutions
Similarly trying other three cases, we get no more solutions, Hence this is the solution for
Finally,
So,
- Arnav Nigam
Solution 4 (No Casework: Two Variables, Two Equations)
Number the given equations and
in this order. Rearranging
and solving for
we have
Substituting
into
and solving for
we get
Substituting
and
into
and simplifying, we rewrite the left side of
in terms of
and
only:
SOLUTION IN PROGRESS. NO EDIT PLEASE--A MILLION THANKS!
~MRENTHUSIASM
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.